How to use querySelectorAll only for elements that have a specific attribute set?

Question

You can use querySelectorAll() like this:

var test = document.querySelectorAll('input[value][type="checkbox"]:not([value=""])');

This translates to:

get all inputs with the attribute "value" and has the attribute "value" that is not blank.

In this demo, it disables the checkbox with a non-blank value.

With your example:

<input type="checkbox" id="c2" name="c2" value="DE039230952"/>

Replace $$ with document.querySelectorAll in the examples:

$$('input') //Every input
$$('[id]') //Every element with id
$$('[id="c2"]') //Every element with id="c2"
$$('input,[id]') //Every input + every element with id
$$('input[id]') //Every input including id
$$('input[id="c2"]') //Every input including id="c2"
$$('input#c2') //Every input including id="c2" (same as above)
$$('input#c2[value="DE039230952"]') //Every input including id="c2" and value="DE039230952"
$$('input#c2[value^="DE039"]') //Every input including id="c2" and value has content starting with DE039
$$('input#c2[value$="0952"]') //Every input including id="c2" and value has content ending with 0952
$$('input#c2[value*="39230"]') //Every input including id="c2" and value has content including 39230

Use the examples directly with:

const $$ = document.querySelectorAll.bind(document);

Some additions:

$$(.) //The same as $([class])
$$(div > input) //div is parent tag to input
document.querySelector() //equals to $$()[0] or $()

Extra Tips:

Multiple "nots", input that is NOT hidden and NOT disabled:

:not([type="hidden"]):not([disabled])

Also did you know you can do this:

node.parentNode.querySelectorAll('div');

This is equivelent to jQuery's:

$(node).parent().find('div');

Which will effectively find all divs in "node" and below recursively, HOT DAMN!